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Unformatted text preview: Lecture 23: Ideal Gas Law and The First Law of Thermodynamics 1 (REVIEW) Chapter 17: Heat Transfer Origin of the calorie unit A few hundred years ago when people were investigating heat and temperature phenomena, the idea was that heat was some kind of invisible caloric fluid which could flow from one material to another. This is the origin of the word calorie, which we now know as a form of energy. A calorie can be defined as the amount of heat energy needed to raise the temperature of water by one Celsius degree, or one Kelvin. 1 Since the heat capacity formula for raising the temperature of a mass is given by Q = mc T ( Heat capacity c is also called specific heat ) this means that the heat capacity number c for water is 1 calorie/gram-K. Methods of heat transfer Heat can be transferred by three methods 1) Conduction. If we have a material of cross-sectional area A and length L , with one end of the length at a high temperature T H and the other length at a low temperature T L , then heat will flow from the high temperature side to the lower temperature side according to: H = dQ dt = kA T H- T L L Here k is the thermal conductivity of the material in units of W/m-K. 2) Convection. Convection involves masses with high temperature moving into regions of low temperature. Rising currents of air from the Earths surface are examples of heat transfer by convection. There is no simple equation for convection as there is for conduction. 3) Radiation. Radiation is the emission of heat energy in the form of electro- magnetic waves from their surface. All objects above absolute zero temper- ature will radiate according to the Stefan-Boltzmann equation H = AeT 4 where A is the area of the bodys surface, e is the emissivity parameter which is dimensionless and is 0 &lt; e 1, is the Stefan-Boltzmann constant with a value = 5 . 67 10- 8 W/m 2-K 4 , and T is temperature in the Kelvin scale. 1 The modern language convention is to simply say Kelvin, and not Kelvin degree or degrees Kelvin. Lecture 23: Ideal Gas Law and The First Law of Thermodynamics 2 The Mechanical Equivalent of Heat Heat energy stored in food is counted in kilocalories or Calories (upper case !) where 1 Calorie = 4,186 Joules. Example A student eats a dinner rated a 2,000 Calories, and the student decides to burn that off by constantly lifting a 50 kg mass (110 lbs) 2 meters high. How many times would the student have to lift such a mass in order to burn off all that dinner? Solution The amount of energy contained in the dinner is given by 2000 4 , 186 = 8 . 37 x 10 7 Joules Lifting a 50 kg mass by 2 meters requires an energy E = mgh = 50 9 . 8 2 = 980 Joules So the total number of times to burn off the dinner is n = 8 . 37 x 10 7 980 = 8 , 840 Assuming it takes 5 seconds to lift weight each time, then the student would need 12 hours total to successfully burn off the dinner....
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